hdu 1247 Hat’s Words

Hat’s Words

　　　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　Total Submission(s): 2414    Accepted Submission(s): 880

Problem Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

 

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

 

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

 

Sample Input

　　a

　　ahat

　　hat

　　hatword

　　hziee

　　word

 

Sample Output

　　ahat

　　hatword

 

题意很清楚:就是给出若干个单词，找出其中能够由其它两个单词连接组合而成的单词。比如上面sample中的"ahat"能够由"a"和"hat"拼凑而成，"hatword"="hat"+"word".因此只需对每个单词进行切分，把切分得到的两个单词在原单词序列中查找，若能查到则符合条件。在这里为了降低查找的复杂度，采用Trie树存储原始单词序列。

实现代码:

/*hdu 1247 Hat's word 字典树 2011.10.16*/ 
#include <iostream>
#include<cstring>
#define MAX 26
using namespace std;

typedef struct Trie_Node
{
	bool isWord;
	struct Trie_Node *next[MAX];
}Trie;

char s[50000][50];

void insert(Trie *root,char *word)      //插入单词 
{
	Trie *p=root;
	while(*word!='\0')
	{
		if(p->next[*word-'a']==NULL)
		{
			Trie *temp=(Trie *)malloc(sizeof(Trie));
			for(int i=0;i<MAX;i++)
			{
				temp->next[i]=NULL;
			}
			temp->isWord=false;
			p->next[*word-'a']=temp;
		}
		p=p->next[*word-'a'];
		word++;
	}
	p->isWord=true;
}

bool search(Trie *root,char *word)         //查找单词是否存在 
{
	Trie *p=root;
	for(int i=0;word[i]!='\0';i++)
	{
		if(p==NULL||p->next[word[i]-'a']==NULL)
			return false;
		p=p->next[word[i]-'a'];
	}
	return p->isWord;
}

void del(Trie *root)                    //释放空间 
{
	for(int i=0;i<MAX;i++)
	{
		if(root->next[i]!=NULL)
		{
			del(root->next[i]);
		}
	}
	free(root);
}


int main(int argc, char *argv[])
{
	int i,j;
	int count=0;
	char str[50];
	Trie *root=(Trie *)malloc(sizeof(Trie));
	for(i=0;i<MAX;i++)
	{
		root->next[i]=NULL;
	}
	root->isWord=false;
	while(scanf("%s",str)!=EOF)
	{	
		strcpy(s[count++],str);
		insert(root,str);	
	}
	for(i=0;i<count;i++)
	{	
		for(j=1;j<=strlen(s[i])-1;j++)	
		{
			char temp1[50]={'\0'};
			char temp2[50]={'\0'};
			strncpy(temp1,s[i],j);     
			strncpy(temp2,s[i]+j,strlen(s[i])-j);
			if(search(root,temp1)&&search(root,temp2))
			{
				printf("%s\n",s[i]);
				break;                       //注意查找成功之后，需跳出循环，否则可能会重复打印
			}
		}
	}
	del(root);
	return 0;
}